Where $f(n)=\dfrac{2}{8n^2-5}$ and $n\in \mathbb{N}$, $\varepsilon\in\ \mathbb{R}^{+}$, $N:\mathbb{R}^{+}\rightarrow \mathbb{N}$. I realise that this is related to epsilon-delta limit proofs but I can't figure out how to get the answer. I have attempted this by just solving for $f(n)=\varepsilon$, however I bieleve that I need to solve it using epsilon delta proof which I have no idea how to do. I've researched the epsilon-delta limit proofs as an attempt to understand the problem but all the example problems give you the limit and make you prove, whereas this seems very different. Sorry for my bad English.

  • $\begingroup$ Let $\dfrac{2}{8n^2-5}<\varepsilon$ and find $n$. $\endgroup$ – Nosrati Jul 31 '18 at 4:27

Solving $f(N)=\epsilon$ for $N$ is indeed a valid way, with some care. More precisely, you should solve $|f(N)-L|=\epsilon$ in case the limit is $L$ (here $l=0$) and the function $f$ might alternate around $L$.

But it is important to ensure that $|f(N)-L|=\epsilon\implies |f(n)-L|<\epsilon$ for all $n>N$, because the difference $|f(N)-L|$ might have larger solutions or even not be decreasing.

Also, in most circumstances, solving exactly is overkill and it is simpler to replace $|f(N)-L|$ by an upper bound, which makes the problem more tractable.

In the case on hand, $L$ is $0$ and for $n>0$ there is no need to take the absolute value. So

$$\frac2{8n^2-5}=\epsilon\iff n=\pm\sqrt{\frac1{4\epsilon}+\frac58}.$$ Of course you take the positive solution, and as the function is decreasing,

$$N=\left\lceil\sqrt{\frac1{4\epsilon}+\frac58}\right\rceil$$ is a possibility.

But isn't it simpler to write (borrowed from Fred)

$$f(N)=\frac2{8N^2-5}<\frac1N<\epsilon$$ that is fulfilled with $N=\left\lceil\dfrac1\epsilon\right\rceil$ ?


Let $n \ge 3$.

$f(n)= \dfrac{2}{8n^2-5} \lt$

$\dfrac{2}{8n^2-n^2}\lt \dfrac{2}{2n^2} \lt 1/n.$

Let $\epsilon \gt 0$ be given.

Archimedean principle:

There is a $n_0$, positive integer, such that

$n_0> 1/\epsilon$.

For $n \ge n_0$ :

$f(n) \lt 1/n \le 1/n_0 \lt \epsilon$.


We have $0 <f(n) \le \frac{1}{n^2}\le \frac{1}{n}$ for all $n$.

Can you proceed ?

  • $\begingroup$ You may disagree, but I sincerely believe that short hints posted with a view of making the asker to see the light should only be comments. $\endgroup$ – Jyrki Lahtonen Jul 31 '18 at 7:17
  • $\begingroup$ Yes, I diagree ! $\endgroup$ – Fred Jul 31 '18 at 7:24

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